Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 16 - A Macroscopic Description of Matter - Exercises and Problems - Page 465: 40


$P = 1.02\times 10^{-16}~atm$

Work Step by Step

We can find the pressure as: $PV = NkT$ $P = (\frac{N}{V})~(kT)$ $P = (\frac{100}{1~cm^3})(\frac{10^6~cm^3}{1~m^3})(1.38\times 10^{-23}~J/K)(7500~K)$ $P = 1.035\times 10^{-11}~Pa$ We then convert the pressure to units of atm: $P = (1.035\times 10^{-11}~Pa)(\frac{1~atm}{1.013\times 10^5~Pa})$ $P = 1.02\times 10^{-16}~atm$
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